Math, asked by nidhanajmaranjith35, 1 year ago

Sinx=1/4 , x in 2nd quadrant... find sinx/2 , cosx/2, tanx/2

Answers

Answered by Pitymys
90

Since  x is in the 2nd quadrant,  x/2 is in the first quadrant.

Given  \sin x=\frac{1}{4}  . Now

 \cos x=\pm \sqrt{1-\sin^2 x} \\<br />\cos x=\pm \sqrt{1-\frac{1}{4}^2}  \\<br />\cos x=\pm \frac{\sqrt{15}}{4}

Since  x is in the 2nd quadrant,  \cos x=- \frac{\sqrt{15}}{4}  &lt;0

Now use the identity,

 2\cos ^2(x/2)=1+\cos x\\<br />2\cos ^2(x/2)=1- \frac{\sqrt{15}}{4} \\<br />\cos ^2(x/2)= \frac{4-\sqrt{15}}{8} \\<br />\cos (x/2)=\sqrt{ \frac{4-\sqrt{15}}{8} } \\<br />\sin (x/2)=\sqrt{1-\cos^2 (x/2)}

 \sin (x/2)=\sqrt{1-\frac{4-\sqrt{15}}{8} }\\<br />\sin (x/2)=\sqrt{ \frac{4+\sqrt{15}}{8} } \\<br />\tan (x/2)=\frac{\sin (x/2)}{\sin (x/2)} \\<br />\tan (x/2)=\frac{\sqrt{ \frac{4+\sqrt{15}}{8} } }{\sqrt{ \frac{4-\sqrt{15}}{8} } } \\<br />\tan (x/2)=\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}}

Answered by vinkalver
2

Step-by-step explanation:

Since xx is in the 2nd quadrant, x/2x/2 is in the first quadrant.

Given \sin x=\frac{1}{4}sinx=

4

1

. Now

\begin{gathered}\cos x=\pm \sqrt{1-\sin^2 x} \\ \cos x=\pm \sqrt{1-\frac{1}{4}^2} \\ \cos x=\pm \frac{\sqrt{15}}{4}\end{gathered}

cosx=±

1−sin

2

x

cosx=±

1−

4

1

2

cosx=±

4

15

Since xx is in the 2nd quadrant, \cos x=- \frac{\sqrt{15}}{4} < 0cosx=−

4

15

<0

Now use the identity,

\begin{gathered}2\cos ^2(x/2)=1+\cos x\\ 2\cos ^2(x/2)=1- \frac{\sqrt{15}}{4} \\ \cos ^2(x/2)= \frac{4-\sqrt{15}}{8} \\ \cos (x/2)=\sqrt{ \frac{4-\sqrt{15}}{8} } \\ \sin (x/2)=\sqrt{1-\cos^2 (x/2)}\end{gathered}

2cos

2

(x/2)=1+cosx

2cos

2

(x/2)=1−

4

15

cos

2

(x/2)=

8

4−

15

cos(x/2)=

8

4−

15

sin(x/2)=

1−cos

2

(x/2)

\begin{gathered}\sin (x/2)=\sqrt{1-\frac{4-\sqrt{15}}{8} }\\ \sin (x/2)=\sqrt{ \frac{4+\sqrt{15}}{8} } \\ \tan (x/2)=\frac{\sin (x/2)}{\sin (x/2)} \\ \tan (x/2)=\frac{\sqrt{ \frac{4+\sqrt{15}}{8} } }{\sqrt{ \frac{4-\sqrt{15}}{8} } } \\ \tan (x/2)=\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}}\end{gathered}

sin(x/2)=

1−

8

4−

15

sin(x/2)=

8

4+

15

tan(x/2)=

sin(x/2)

sin(x/2)

tan(x/2)=

8

4−

15

8

4+

15

tan(x/2)=

4−

15

4+

15

Similar questions